Question

A
ground state hydrogen atom absorbs a photon light having a
wavelength of 92.57 nm. It then gives off a photon having a
wavelength of 1944 nm. What is the final state of the hydrogen
atom?

I got Nf=1.14 and it wasn't right

Answer #1

Absorbed photon energy Ea = hc/λ.. (Planck's equation) =Ea = hc
/ 92.57x10^{-9}m

Energy emitted .. Ee = hc/ 1944x10^{-9}m

Energy retained .. ∆E = Ea - Ee =hc(1/92.57^-9 -
1/1944^-9)

∆E = hc(10^{9}/92.57-10^{9}/1944)

=hcx10^{9}(0.0108-0.01.286)

∆E
=(6.625x10^{-34})(3x10^{8})x10^{7}x1.0286

∆E =20.40x10^{-19}

∆E =2.040x10^{-18}J

converting J to eV

∆E =2.040x10^{-18}/1.60x10^{-19}

∆E =12.75eV

New energy state = (-13.60 + 12.75)eV = -0.85 eV

Energy states for Hydrogen .. En = - (13.60 / n²)

n² = -13.60 / -0.85 = 16 .. .. .. ►n = 4

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please try to show solution

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